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\begin{document}
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\title{Numerical Analysis homework \# 3}

\author{王劼 Wang Jie 3220100105
  \thanks{Electronic address: \texttt{2645443470@qq.com}}}
\affil{(math), Zhejiang University }


\date{Due time: \today}

\maketitle

\begin{abstract}
    theoretical homework 3  
\end{abstract}





% ============================================
\section*{theoretical homework}

Complete all the exercises in section 3.4.1. You are strongly encouraged to use LaTeX, or else you should make sure that your handwriting is recognizable, otherwise you only get partial credit for the recognizable part. In addition, please upload your pdf file format: HW3\_YourStuNum\_YourName.pdf only. \cite{wangheyu2024}

\subsection*{Question 3.4.1 I}  

\textbf{Answer:} \\

To determine \( p \in P_3 \) such that \( s(0) = 0 \), we need to find a cubic polynomial \( p(x) \) that equals zero when \( x = 0 \). Since \( s(x) \) is defined by \( p(x) \) for \( x \in [0,1] \), we require \( p(0) = 0 \).

Let \( p(x) = ax^3 + bx^2 + cx \). Given \( p(0) = 0 \), we have \( d = 0 \). Thus, \( p(x) = ax^3 + bx^2 + cx \).

Next, we must consider the continuity and smoothness of \( s(x) \) at \( x = 1 \) to ensure that \( s(x) \in S^2_3\), the function value, first derivative, and second derivative are continuous at each internal node.

1. Function value continuity: \( p(1) = (2-1)^3 \)
2. First derivative continuity: \( p'(1) = -3(2-1)^2 \)
3. Second derivative continuity: \( p''(1) = 6(2-1) \)

Since \( s(x) \) is defined by \( (2-x)^3 \) at \( x = 1 \), we have:
- \( (2-1)^3 = 1 \)
- \( -3(2-1)^2 = -3 \)
- \( 6(2-1) = 6 \)

We need to solve the following system of equations:
\begin{align*}
a + b + c &= 1 \\
3a + 2b + c &= -3 \\
6a + 2b &= 6
\end{align*}

Solving this system, we get:
\begin{align*}
   a &= 7 \\
   b &= -18 \\
   c &= 12
\end{align*}

Thus, \( p(x) = 7x^3 - 18x^2 + 12x \).

Finally, we need to check if \( s(x) \) is a natural cubic spline. Since the second derivative of \( s(x) \) at \( x = 0 \) is -36. 

So \( s(x) \) is not a natural cubic spline.

\subsection*{Question 3.4.1 II} 

\textbf{Answer:} \\

Given $f_i = f(x_i)$ of some scalar function at points $a = x_1 < x_2 < \cdots < x_n = b$, we consider interpolating $f$ on $[a, b]$ with a quadratic spline $s \in S_2^1$.

\subsection*{(a) Why is an additional condition needed to determine $s$ uniquely?}

In quadratic spline interpolation, each subinterval $[x_i, x_{i+1}]$ has a quadratic polynomial as the spline function $s$. 
For $n-1$ subintervals, we have $3(n-1)$ coefficients to determine. 
However, we only have $n$ function values $f_i$ as conditions, providing $n$ equations. 
Additionally, for continuity and first derivative continuity each internal node $x_i$.
This provides $2(n-2)$ more conditions for each internal node. 
So total we get $3n-4$ conditions as equations, and we have $3n-3$ coefficients to determine.
Thus, we need an additional condition to determine $s$ uniquely.

\subsection*{(b) Define $m_i = s'(x_i)$ and $p_i = s|_{[x_i, x_{i+1}]}$. Determine $p_i$ in terms of $f_i, f_{i+1}$, and $m_i$ for $i = 1, 2, \ldots, n-1$.}

The quadratic spline $p_i$ on the interval $[x_i, x_{i+1}]$ can be expressed by n the Newton formula as:
\[ p_i(x) = f_i + m_i(x-x_i) + f[x_i,x_i,x_{i+1}](x-x_i)^2 \]
where $f[x_i,x_i,x_{i+1}] = \frac{f[x_i,x_{i+1}]-m_i}{x_{i+1}-x_i} = \frac{f_{i+1}-f_i-m_i(x_{i+1}-x_i)}{(x_{i+1}-x_i)^2}$ are coefficients to be determined. Using the conditions $p_i(x_i) = f_i$, $p_i(x_{i+1}) = f_{i+1}$, and $p_i'(x_i) = m_i$, $p_i'(x_{i+1}) = m_{i+1}$.
We get:
\[ p_i(x) = f_i + m_i(x-x_i) + f[x_i,x_i,x_{i+1}](x-x_i)^2 = f_i + m_i(x-x_i) + \frac{f_{i+1}-f_i-m_i(x_{i+1}-x_i)}{(x_{i+1}-x_i)^2}(x-x_i)^2 \]

\subsection*{(c) Suppose $m_1 = f'(a)$ is given. Show how $m_2, m_3, \ldots, m_{n-1}$ can be computed.}

If $m_1$ is known, we can use the condition that first derivative continuity of the spline function at the internal nodes to compute $m_2, m_3, \ldots, m_{n-1}$. For each internal node $x_i$, we have:
\begin{align*}
   p_i'(x_{i+1}) = p_{i+1}'(x_{i+1}) \\
   m_i + 2(x_{i+1}-x_i)^2f[x_i,x_i,x_{i+1}] &= m_{i+1} + 2(x_{i+1}-x_i)^2f[x_{i+1},x_{i+1},x_{i+2}] \\
   m_i + 2(x_{i+1}-x_i)^2f[x_i,x_i,x_{i+1}] - 2(x_{i+1}-x_i)^2f[x_{i+1},x_{i+1},x_{i+2}] &= m_{i+1}
\end{align*}

Thus, we arrive at the recursive formulas.

Because $m_1 = f'(a)$ is given, the recursive formulas implies $m_2, m_3, \ldots, m_{n-1}$ can be computed.

\subsection*{Question 3.4.1 III} 

\textbf{Answer:}\\ 
Let \( s_1(x) = 1 + c(x + 1)^3 \) where \( x \in [-1, 0] \) and \( c \in \mathbb{R} \). Determine \( s_2(x) \) on \([0, 1]\) such that
\[ s(x) = \begin{cases} 
s_1(x) & \text{if } x \in [-1, 0], \\
s_2(x) & \text{if } x \in [0, 1]
\end{cases} \]
is a natural cubic spline on \([-1,1]\) with knots \(-1, 0, 1\). How must \( c \) be chosen if one wants \( s(1) = -1 \)?

First, we know that \( s_1(x) \) on \([-1, 0]\) is given by \( s_1(x) = 1 + c(x + 1)^3 \). We need to find \( s_2(x) \) on \([0, 1]\).

We can assume \( s_2(x) \) has the form \( s_2(x) = Ax^3 + Bx^2 + Cx + D \). We need to find \( A, B, C, D \) such that:

1. \( s_2(0) = s_1(0) \)
2. \( s_2'(0) = s_1'(0) \)
3. \( s_2''(0) = s_1''(0) \)
4. \( s_2''(1) = 0 \)

Calculating \( s_1(x) \) at \( x = 0 \) and its derivatives:
- \( s_1(0) = 1 + c(0 + 1)^3 = 1 + c \)
- \( s_1'(x) = 3c(x + 1)^2 \), so \( s_1'(0) = 3c \)
- \( s_1''(x) = 6c(x + 1) \), so \( s_1''(0) = 6c \)

We have the following system of equations:
\begin{align*}
   A &= -c \\
   B &= 3c \\
   C &= 3c \\
   D &= 1+c 
\end{align*}

So we get \( s_2(x) = -cx^3 + 3cx^2 + 3cx + 1 + c \).

When \( s(1)=-1 \), we have $c=-\frac{1}{3}$

\subsection*{Question 3.4.1 IV}

\textbf{Answer:}\\

Consider \( f(x) = \cos\left(\frac{\pi}{2}x\right) \) with \( x \in [-1, 1] \).

\subsection*{(a) Determine the natural cubic spline interpolant to \( f \) on knots -1, 0, 1.}

First, we calculate the values of \( f(x) \) at the knots:
\begin{align*}
f(-1) &= \cos\left(-\frac{\pi}{2}\right) = 0, \\
f(0) &= \cos(0) = 1, \\
f(1) &= \cos\left(\frac{\pi}{2}\right) = 0.
\end{align*}

The natural cubic spline \( s(x) \) can be expressed as:
\[ s(x) = \begin{cases} 
p_1(x) & \text{if } x \in [-1, 0], \\
p_2(x) & \text{if } x \in [0, 1]
\end{cases} \]

where \( p_1(x) \) and \( p_2(x) \) are quadratic polynomials. Since it is a natural cubic spline, we have:
\[ p_1''(-1) = 0 \]
\[ p_2''(1) = 0 \]

For $M_i=s''(x_i)$ and $m_i=s'(x_i)$ when $x_1=-1,x_2=0,x_3=1$, we get:
\begin{align*}
M_1 &= 0, \\
M_2 &= -3, \\
M_3 &= 0, \\
p_1(x) &= \frac{3(x+1)}{2}-\frac{(x+1)^3}{2}, \\
p_2(x) &= 1-\frac{3x^2}{2}-\frac{x^3}{2}.
\end{align*}

So the natural cubic spline \( s(x) \) can be expressed as:
\[ s(x) = \begin{cases} 
\frac{3(x+1)}{2}-\frac{(x+1)^3}{2} & \text{if } x \in [-1, 0], \\
1-\frac{3x^2}{2}+\frac{x^3}{2} & \text{if } x \in [0, 1]
\end{cases} \]

\subsection*{(b) Verify that natural cubic splines have the minimal total bending energy.}

To verify that natural cubic splines have the minimal total bending energy, we compare the bending energy of the natural cubic spline and \( g(x) \). The bending energy \( E \) is defined as:
\[ E = \int_a^b (s''(x))^2 \, dx \]

\subsection*{(i)the quadratic polynomial that interpolates f at -1, 0, 1}

For the quadratic polynomial $g(x)$ that interpolates f at -1, 0, 1, when $m_1=f'(-1)$ we can get :
\[ g(x) = \begin{cases} 
m_1(x+1)+(1-m_1)(x+1)^2 & \text{if } x \in [-1, 0], \\
1+(2-m_1)x+(m_1-3)x^2 & \text{if } x \in [0, 1]
\end{cases} \]

\[ g''(x) = \begin{cases} 
2-2m_1 & \text{if } x \in [-1, 0], \\
2m_1-6 & \text{if } x \in [0, 1]
\end{cases} \]

\[ \int_{-1}^1 (s''(x))^2 \, dx = 6, \text{while} \int_{-1}^1 (g''(x))^2 \, dx = 8(m_1^2-4m_1+5) \geq 8\]

Thus, we have \(\int_{-1}^1 (s''(x))^2 \, dx \leq \int_{-1}^1 (g''(x))^2 \, dx\).

\subsection*{(ii)f(x)}

Consider $g(x) = f(x) = \cos\left(\frac{\pi}{2}x\right)$, we get:

\[\int_{-1}^1 (g''(x))^2 \, dx = \frac{\pi^4}{16} \geq 6 = \int_{-1}^1 (s''(x))^2 \, dx\]

So we konw natural cubic splines have the minimal total bending energy.

\subsection*{Question 3.4.1 V}

\textbf{Answer:}\\

\subsection*{(a) Derive the explicit expression of \( B_i^2(x) \)}
The quadratic B-spline \( B_i^2(x) \) can be derived from the recursive definition of B-splines and the hat function. The recursive definition is given by:
\[ B_i^1(x) = \hat{B_i}(x) = \begin{cases} 
\frac{x-t_{i-1}}{t_i-t_{i-1}} & \text{if } x \in (t_{i-1},t_i] \\
\frac{t_{i+1}-x}{t_{i+1}-t_i} & \text{if } x \in (t_i,t_{i+1}] \\
0 & \text{otherwise }
\end{cases} \]

\[ B_i^{k+1}(x) = \frac{x - t_i}{t_{i+k+1} - t_i} B_i^k(x) + \frac{t_{i+k+2} - x}{t_{i+k+2} - t_{i+1}} B_{i+1}^k(x) \]

For the quadratic B-spline \( B_i^2(x) \), we start from \( B_i^1(x) \) and apply the recursive formula.

So we get:
\[ B_i^2(x) = \begin{cases} 
\frac{(x-t_{i-1})^2}{(t_{i+1}-t_{i-1})(t_i-t_{i-1})} & \text{if } x \in (t_{i-1},t_i] \\
\frac{(x-t_{i-1})(t_{i+1}-x)}{(t_{i+1}-t_{i-1})(t_{i+1}-t_i)}+\frac{(x-t_{i})(t_{i+2}-x)}{(t_{i+2}-t_i)(t_{i+1}-t_i)} & \text{if } x \in (t_i,t_{i+1}] \\
\frac{(x-t_{i+2})^2}{(t_{i+2}-t_i)(t_{i+2}-t_{i+1})} & \text{if } x \in (t_{i+1},t_{i+2}] \\
0 & \text{otherwise }
\end{cases} \]

\subsection*{(b) Verify the first derivate continuity of \( B_i^2(x) \) at \( t_i \) and \( t_{i+1} \)}
To verify the first derivate continuity of \( B_i^2(x) \) at \( t_i \) and \( t_{i+1} \).

\[\frac{d}{dx} B_i^2(x) = \begin{cases} 
   \frac{2(x-t_{i-1})}{(t_{i+1}-t_{i-1})(t_i-t_{i-1})} & \text{if } x \in (t_{i-1},t_i] \\
   \frac{t_{i+1}+t_{i-1}-2x}{(t_{i+1}-t_{i-1})(t_{i+1}-t_i)}+\frac{t_{i+2}+t_i-2x}{(t_{i+2}-t_i)(t_{i+1}-t_i)} & \text{if } x \in (t_i,t_{i+1}] \\
   \frac{2(x-t_{i+2})}{(t_{i+2}-t_i)(t_{i+2}-t_{i+1})} & \text{if } x \in (t_{i+1},t_{i+2}] \\
   0 & \text{otherwise }
   \end{cases} \]

For $x=t_i,t_{i+1}$, from the expression of $\frac{d}{dx} B_i^2(x)$, we get the continuity of \( B_i^2(x) \) at \( t_i \) and \( t_{i+1} \).

\subsection*{(c) Show the existence of a unique root in \( (t_{i-1}, t_{i+1}) \)}
From the expression of $\frac{d}{dx} B_i^2(x)$, we get :
\[\frac{d}{dx} B_i^2(x) = 0, \text{ which equals to} \frac{t_{i+1}+t_{i-1}-2x}{(t_{i+1}-t_{i-1})(t_{i+1}-t_i)}+\frac{t_{i+2}+t_i-2x}{(t_{i+2}-t_i)(t_{i+1}-t_i)} = 0, x \in (t_i,t_{i+1}] \]

so there is only one solution $x^* = \frac{t_{i+1}t_{i+2}-t_{i-1}t_i}{t_{i+1}+t_{i+2}-t_{i-1}-t_i}$

\subsection*{(d) Show \( B_i^2(x) \in [0, 1) \)}
Since \( B_i^2(x) \) is derived from the hat function and is a quadratic polynomial, the hat function achieves its maximum value at \( t_i \), which is 1. 

\[ B_i^2(x) = \frac{x - t_{i-1}}{t_{i+1} - t_{i-1}} \hat{B_i}(x) + \frac{t_{i+2} - x}{t_{i+2} - t_i} \hat{B_{i+1}}(x) \]

We can find the coefficients are less than 1 within the the interval of support, such as \(0 \leq \frac{x - t_{i-1}}{t_{i+1} - t_{i-1}} \leq 1\) within the interval of $\hat{B_i}(x)$ support, which is $(t_{i-1},t{i+1}]$.

So \( B_i^2(x) \in [0, 1) \)

\subsection*{(e) Plot \( B_i^2(x) \) for \( t_i = i \)}
The picture shows:

\begin{figure}[htbp]
   \centering
   \includegraphics[width=0.7\textwidth]{b_splines_plot.png}  
   \caption{\(B_i^2(x) \) for \( t_i = i \)}
   \label{b_splines_plot}
\end{figure}

\subsection*{Question 3.4.1 VI}

\textbf{Answer:}\\
We can get:
\begin{align*}
&(t_{i+2} - t_{i-1})[t_{i-1}, t_i, t_{i+1}, t_{i+2}](t-x)_+^2 \\
=& [t_i, t_{i+1}, t_{i+2}](t-x)_+^2 - [t_{i-1}, t_i, t_{i+1}](t-x)_+^2 \\
=& \frac{1}{t_{i+2} - t_i} ([t_{i+1}, t_{i+2}](t-x)_+^2 - [t_i, t_{i+1}](t-x)_+^2) \\
& - \frac{1}{t_{i+1} - t_{i-1}} ([t_i, t_{i+1}](t-x)_+^2 - [t_{i-1}, t_i](t-x)_+^2) \\
=& \frac{1}{t_{i+2} - t_i} \left[ \frac{(t_{i+2} - x)_+^2 - (t_{i+1} - x)_+^2}{t_{i+2} - t_{i+1}} - \frac{(t_{i+1} - x)_+^2 - (t_i - x)_+^2}{t_{i+1} - t_i} \right] \\
& - \frac{1}{t_{i+1} - t_{i-1}} \left[ \frac{(t_{i+1} - x)_+^2 - (t_i - x)_+^2}{t_{i+1} - t_i} - \frac{(t_i - x)_+^2 - (t_{i-1} - x)_+^2}{t_i - t_{i-1}} \right] \\
=& \begin{cases} 
\frac{(x-t_{i-1})^2}{(t_{i+1}-t_{i-1})(t_i-t_{i-1})} & \text{if } x \in (t_{i-1},t_i] \\
\frac{(x-t_{i-1})(t_{i+1}-x)}{(t_{i+1}-t_{i-1})(t_{i+1}-t_i)}+\frac{(x-t_{i})(t_{i+2}-x)}{(t_{i+2}-t_i)(t_{i+1}-t_i)} & \text{if } x \in (t_i,t_{i+1}] \\
\frac{(x-t_{i+2})^2}{(t_{i+2}-t_i)(t_{i+2}-t_{i+1})} & \text{if } x \in (t_{i+1},t_{i+2}] \\
0 & \text{otherwise } 
\end{cases} \\
=& B_i^2
\end{align*}

\subsection*{Question 3.4.1 VII}

\textbf{Answer:}\\

We want to prove the following by induction: 
\[
\frac{\int_{t_{i-1}}^{t_{i+n}} B_i^n(x) \, dx}{t_{i+n} - t_{i-1}} = \frac{1}{n+1}.
\]

\textbf{Base Case}: \( n = 1 \)

For \( n = 1 \), the Bernstein polynomial is simply:
\[ B_i^1(x) = \hat{B_i}(x) = \begin{cases} 
\frac{x-t_{i-1}}{t_i-t_{i-1}} & \text{if } x \in (t_{i-1},t_i] \\
\frac{t_{i+1}-x}{t_{i+1}-t_i} & \text{if } x \in (t_i,t_{i+1}] \\
0 & \text{otherwise }
\end{cases} \]
and it is easy to verify that:
\[
\frac{\int_{t_{i-1}}^{t_{i+1}} B_i^1(x) \, dx}{t_i - t_{i-1}} = \frac{1}{2}.
\]
So the base case holds.

\textbf{Inductive Step}

Now, assume the result holds for n-1. That is, we assume:
\[
\frac{\int_{t_{i-1}}^{t_{i+n-1}} B_i^{n-1}(x) \, dx}{t_{i+n-1} - t_{i-1}} = \frac{1}{n}.
\]
We need to prove that:
\[
\frac{\int_{t_{i-1}}^{t_{i+n}} B_i^n(x) \, dx}{t_{i+n} - t_{i-1}} = \frac{1}{n+1}.
\]

We know the recursive formula for \( B_i^n(x) \) is:
\[
B_i^n(x) = \frac{x - t_{i-1}}{t_{i+n-1} - t_{i-1}} B_i^{n-1}(x) + \frac{t_{i+n} - x}{t_{i+n} - t_i} B_{i+1}^{n-1}(x),
\]
and its derivative with respect to \( x \) is:
\[
\frac{d B_i^n(x)}{d x} = \frac{n B_i^{n-1}(x)}{t_{i+n-1} - t_{i-1}} - \frac{n B_{i+1}^{n-1}(x)}{t_{i+n} - t_i}.
\]

Using this, we begin with the integral:
\[
\int_{t_{i-1}}^{t_{i+n}} B_i^n(x) \, dx = - \int_{t_{i-1}}^{t_{i+n}} x \, dB_i^n(x).
\]
Substitute for \( dB_i^n(x) \):
\[
= - \int_{t_{i-1}}^{t_{i+n}} x \left( \frac{n B_i^{n-1}(x)}{t_{i+n-1} - t_{i-1}} - \frac{n B_{i+1}^{n-1}(x)}{t_{i+n} - t_i} \right) dx.
\]

We can also conclude from the recursive formula of \( B_i^n(x) \) that:
\[
\int_{t_{i-1}}^{t_{i+n}} B_i^n(x) \, dx = \int_{t_{i-1}}^{t_{i+n}} \frac{x B_i^{n-1}(x)}{t_{i+n-1} - t_{i-1}} \, dx - \int_{t_{i-1}}^{t_{i+n}} \frac{x B_{i+1}^{n-1}(x)}{t_{i+n} - t_i} \, dx + \int_{t_{i-1}}^{t_{i+n}} \frac{B_{i+1}^{n-1}(x) t_{i+n}}{t_{i+n} - t_i} \, dx - \int_{t_{i-1}}^{t_{i+n}} \frac{B_i^{n-1}(x) t_{i-1}}{t_{i+n-1} - t_{i-1}} \, dx.
\]

Now, from the above equations we rearrange and compute the final integral:
\[
(n+1) \int_{t_{i-1}}^{t_{i+n}} B_i^n(x) \, dx = - \int_{t_{i-1}}^{t_{i+n}} \left( \frac{n B_i^{n-1}(x) t_{i-1}}{t_{i+n-1} - t_{i-1}} - \frac{n B_{i+1}^{n-1}(x) t_{i+n}}{t_{i+n} - t_i} \right) dx.
\]

At this point, simplifying the integrals yields:
\[
= t_{i+n} - t_{i-1}.
\]

Thus, we have the final result:
\[
\frac{\int_{t_{i-1}}^{t_{i+n}} B_i^n(x) \, dx}{t_{i+n} - t_{i-1}} = \frac{1}{n+1}.
\]

Therefore, the inductive step holds, and by induction, the result is true for all \( n \).

\subsection*{Question 3.4.1 VIII}

\textbf{Answer:}\\

\subsection*{(a) Verify this theorem for m = 4 and n = 2}
To verify \(\tau_{m-n}(x_i, \ldots, x_{i+n}) = [x_i, \ldots, x_{i+n}] x^m\) for m = 4 and n = 2.

\begin{table}[ht]
\centering
\resizebox{\textwidth}{!}{
\begin{tabular}{|c|c|c|c|}
\hline
point & $[x_i], [x_{i+1}], [x_{i+2}]$ & $[x_i, x_{i+1}], [x_{i+1}, x_{i+2}]$ & $[x_i, x_{i+1}, x_{i+2}]$ \\
\hline
$x_i$ & $x_i^4$ & $x_{i+1}^3 + x_{i+1}^2 x_i + x_{i+1} x_i^2 + x_i^3$ & 
$x_{i+2}^2+x_{i+1}^2+x_i^2+x_{i+2}x_{i+1}+x_{i+2}x_i+x_{i+1}x_i$ \\
\hline
$x_{i+1}$ & $x_{i+1}^4$ & $x_{i+2}^3 + x_{i+2}^2 x_{i+1} + x_{i+2} x_{i+1}^2 + x_{i+1}^3$ & \\
\hline
$x_{i+2}$ & $x_{i+2}^4$ & & \\
\hline
\end{tabular}
}
\caption{divided difference table}
\end{table}

So we get:
\[[x_i, x_{i+1}, x_{i+2}] x^4 = x_{i+2}^2+x_{i+1}^2+x_i^2+x_{i+2}x_{i+1}+x_{i+2}x_i+x_{i+1}x_i\]

Thus, we have:
\[\tau_{2}(x_i, x_{i+1}, x_{i+2}) = [x_i, x_{i+1}, x_{i+2}] x^4\]

\subsection*{(b) Prove this theorem by the lemma on the recursive relation on complete symmetric polynomials}
By the recursive relation on complete symmetric polynomials, we have:
\begin{align*}
\tau_{k+1}(x_1, \ldots, x_n, x_{n+1}) &= \tau_{k+1}(x_1, \ldots, x_n) + x_{n+1} \tau_{k}(x_1, \ldots, x_n, x_{n+1}).
\end{align*}

\(\tau_{m-n}(x_i, \ldots, x_{i+n}) = [x_i, \ldots, x_{i+n}] x^m\), we can prove by an induction on \(n\). For \(n=0\), it reduces to
\[
\tau_m(x_i) = [x_i] x^m,
\]
which is trivially true. Now suppose it holds for a non-negative integer \(n < m\). Then the induction hypothesis yield
\begin{align*}
\tau_{m-n-1}(x_i, \ldots, x_{i+n+1}) &= \frac{\tau_{m-n}(x_{i+1}, \ldots, x_{i+n+1}) - \tau_{m-n}(x_i, \ldots, x_{i+n})}{x_{i+n+1} - x_i} \\
&= \frac{[x_{i+1}, \ldots, x_{i+n+1}] x^m - [x_i, \ldots, x_{i+n}] x^m}{x_{i+n+1} - x_i} \\
&= [x_i, \ldots, x_{i+n+1}] x^m,
\end{align*}
which completes the proof.


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\section*{ \center{\normalsize {Acknowledgement}} }
None.


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